Answer to Question #241984 in Physics for Kristine

Question

The position of a squirrel running in a park is given by 𝑟 = [(310.280𝑚/𝑠)𝑡 + (0.0360𝑚/𝑠 2 )𝑡 2 ]𝑖̂+ (0.0190𝑚/𝑠 3 )𝑡 3 𝑗̂. (a) What are 𝑣𝑥(𝑡) and 𝑣𝑦(𝑡), the 𝑥- and 𝑦-components of the velocity of the squirrel, as functions of time? (b) At 𝑡 = 5.00 s, how far is the squirrel from its initial position? (c) At 𝑡 = 5.00 s, what are the magnitude and direction of the squirrel’s velocity?

Answer 1

(a) $v_x(t)=310.280+0.072t$

$v_y(t)=0.057t^2$

(b) $l=\sqrt{(310.280\cdot5+0.036\cdot5^2)^2+(0.019\cdot5^3)^2}=1552.3\ (m)$

(c) $v=\sqrt{(310.280+0.072\cdot5)^2+(0.057\cdot5^2)^2}=311\ (m/s)$

$\tan\alpha=\frac{v_y}{v_x}=\frac{1.425}{310.64}=0.0046\to \alpha=0.26°$ (above the x-axis)